Conventional hydraulic elevators utilize a hydraulically driven ram to raise and lower an elevator car. Conversely, traction elevators utilize a wire rope having one end attached to a car and a second end attached to a counterweight. The rope passes over a sheave which is driven by an electric motor. Traction forces generated between the sheave and the rope raise and lower an elevator car.
Hydraulic elevators require a great deal more installed power compared to conventional traction elevators. The motor in an hydraulic elevator has to deliver the total energy corresponding to the total weight of the car plus the total weight of the load inside the car. The hydraulic elevator may require more than four times the power used by the same size traction elevators since the use of the counterweight balances the weight of the car and approximately 50% of the weight of the load. Basically the required power (Pt) in a traction elevator is: ##EQU1## Where DL=Duty Load
V=Duty Speed PA1 Car Wght=C.sub.W =1.2 D.sub.L PA1 Cwnb=Non balanced car weight PA1 Cwnb=20% D.sub.L
In contrast, in an hydraulic elevator the weight of the car should be a minimum 20% heavier than the duty load to provide the minimum fluid pressure force required by the hydraulic system to operate accurately in an empty car down condition. As such the required power P.sub.H in an hydraulic elevator is: EQU P.sub.H =(C.sub.W +D.sub.L).times.V
Where
And, substituting for C.sub.W and V: EQU P.sub.H =(1.2D.sub.L +D.sub.L).times.V=2.2D.sub.L .times.V=4.4Pt.
Because the required power for the hydraulic elevator is 4.4 times that of the traction elevator, motor power is dimensioned for more than four times the equivalent motor power of a traction elevator.
To address the problem, some hydraulic elevators are equipped with a counterweight which balances approximately 80% of the car weight since a minimum pressure is still needed for an empty car down condition to permit the hydraulic systems to operate accurately. These systems require power as follows: EQU Phcwt=(Cwnb+D.sub.L).times.V.sub.L
Where
And substituting for Cwnb and V.sub.L : EQU Phcwt=(0.2D.sub.L +D.sub.L).times.V.sub.L =1.2D.sub.L V.sub.L =2.4Pt
Such counterweighted hydraulic elevators require a motor to have more than twice the peak power required for an equivalent traction elevator.